Problem 69 - Introductory Chemistry Atoms First

Question

Consider the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) (a) What volume in liters of oxygen gas at STP is required to produce 2 moles of sulfur trioxide gas? (b) What volume in liters of oxygen gas at \(25.0^{\circ} \mathrm{C}\) and 1 atm pressure is required to produce 2 moles of sulfur trioxide gas?

Step-by-Step Solution

Answer

(a) The required volume of oxygen gas at STP to produce 2 moles of sulfur trioxide is \(22.4 \,\mathrm{L}\). (b) The required volume of oxygen gas at \(25.0^{\circ}\mathrm{C}\) and 1 atm pressure to produce 2 moles of sulfur trioxide is \(24.5 \,\mathrm{L}\).

Step 1: Write down the balanced chemical equation

The given balanced chemical equation is: \(2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{SO}_{3}(g)\)

Step 2: Use the stoichiometry to find the moles of oxygen required

According to the balanced equation, 1 mole of oxygen \(\mathrm{O}_{2}\) reacts with 2 moles of sulfur dioxide \(\mathrm{SO}_{2}\) to produce 2 moles of sulfur trioxide \(\mathrm{SO}_{3}\). So, to produce 2 moles of \(\mathrm{SO}_{3}\), we would need 1 mole of \(\mathrm{O}_{2}\).

Step 3a: Find the required volume of oxygen at STP

At STP, the conditions are 273.15 K temperature and 1 atm pressure. Using the ideal gas law formula with n = 1 mole, R = 0.0821 L⋅atm/mol⋅K, T = 273.15 K, and P = 1 atm: \(V = \frac{nRT}{P} = \frac{(1)(0.0821)(273.15)}{1} = 22.4 \,\mathrm{L}\) So, the required volume of oxygen gas at STP to produce 2 moles of sulfur trioxide is 22.4 L.

Step 3b: Find the required volume of oxygen at 25.0°C and 1 atm pressure

Now, we need to find the volume of oxygen gas at 25.0°C (298.15 K) and 1 atm pressure. Using the ideal gas law formula with n = 1 mole, R = 0.0821 L⋅atm/mol⋅K, T = 298.15 K, and P = 1 atm: \(V = \frac{nRT}{P} = \frac{(1)(0.0821)(298.15)}{1} = 24.5 \,\mathrm{L}\) So, the required volume of oxygen gas at 25.0°C and 1 atm pressure to produce 2 moles of sulfur trioxide is 24.5 L.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry used to relate the pressure, volume, temperature, and moles of a gas. The formula is expressed as \(PV = nRT\), where:\

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• \(P\) stands for pressure, usually measured in atmospheres (atm).
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• \(V\) is the volume of the gas in liters (L).
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• \(n\) represents the number of moles of gas.
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• \(R\) is the universal gas constant, approximately 0.0821 L⋅atm/mol⋅K.
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• \(T\) is the absolute temperature in Kelvin (K).
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This equation shows the relationship between a gas’s characteristics under various conditions. If you know three of these variables, you can solve for the fourth. This is particularly useful in situations where you need to determine how a gas behaves under different temperatures or pressures. For example, in the given problem, the ideal gas law helps to calculate the volume of oxygen needed at different temperatures and pressures while producing sulfur trioxide.

Balanced Chemical Equation

In chemistry, a balanced chemical equation is vital for showing the exact proportions in which reactants combine to form products. For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\), the equation depicts how sulfur dioxide and oxygen react to produce sulfur trioxide. \
A chemical reaction involves breaking and forming chemical bonds. Therefore, both sides of the equation must have the same number of each type of atom. This allows us to use the equation to predict the quantities of products formed from given quantities of reactants. In our example, two moles of sulfur dioxide react with one mole of oxygen to yield two moles of sulfur trioxide. Hence, the equation is balanced with respect to all involved elements: sulfur, oxygen.

Reaction Stoichiometry

Reaction stoichiometry involves using a balanced chemical equation to calculate the mass and quantity of reactants and products. This concept is essential for determining how much reactant is required or how much product will be formed in a chemical reaction.
In the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\), stoichiometry informs us that for every mole of oxygen consumed, two moles of sulfur dioxide are used, resulting in two moles of sulfur trioxide.
Using stoichiometry, if you want to produce 2 moles of sulfur trioxide, you'd require exactly 1 mole of oxygen. This provides a clear pathway for calculating the needed volumes and masses in various chemical scenarios, especially when combined with the ideal gas law.

Moles and Volume Calculations

Moles and volume calculations are key in solving gas-related problems, particularly when using the ideal gas law. At standard temperature and pressure (STP), which is 273.15 K and 1 atm, one mole of any ideal gas occupies 22.4 liters.
For example, in part (a) of our original exercise, we calculate the volume of oxygen needed at STP to produce a certain amount of sulfur trioxide. Given that we need 1 mole of \( \mathrm{O}_{2} \) for 2 moles of \( \mathrm{SO}_{3} \), we find that this 1 mole of oxygen gas would occupy 22.4 liters at STP.
In part (b), the conditions change to 25.0°C (298.15 K) at 1 atm. Using the ideal gas law, one mole of oxygen, under these conditions, occupies 24.5 liters. These calculations illustrate how temperature affects gas volume and underscores the importance of considering all parameters when dealing with gas reactions.